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Chapter 28: Problem 56

What is the magnitude of the magnetic field inside a long, straight tungsten wire of circular cross section with diameter \(2.4 \mathrm{~mm}\) and carrying a current of \(3.5 \mathrm{~A}\), at a distance of \(0.60 \mathrm{~mm}\) from its central axis?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field inside the tungsten wire at a distance of 0.60 mm from its central axis is approximately 2.92 x 10^-3 T.

Step by step solution

01

Write down the formula for the magnetic field inside a wire

To calculate the magnetic field at a distance r from the central axis of a long straight wire carrying a current I, we use Ampere's Law formula: \[B = \frac{\mu_{0} I r}{2 \pi R^2}\] where B is the magnetic field, \(\mu_{0}\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{~T \cdot m / A}\)), I is the current, r is the distance from the central axis, and R is the radius of the wire. Keep in mind that the wire has a diameter of \(2.4 \mathrm{~mm}\), so the radius R is half of that: \[R = \frac{2.4 \mathrm{~mm}}{2} = 1.2 \mathrm{~mm} = 1.2 \times 10^{-3} \mathrm{~m}\]
02

Plug in the given values

Now, we can plug in all the given values into the equation and solve for B. The current I is \(3.5 \mathrm{~A}\), and the distance r from the central axis is \(0.60 \mathrm{~mm} = 0.60 \times 10^{-3} \mathrm{~m}\): \[B = \frac{4\pi \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 3.5 \mathrm{~A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{2 \pi \cdot (1.2 \times 10^{-3} \mathrm{~m})^2}\]
03

Calculate the magnetic field magnitude

Now, simplify the expression and calculate the magnetic field magnitude: \[B = \frac{4\pi \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 3.5 \mathrm{~A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{2 \pi \cdot 1.44 \times 10^{-6} \mathrm{~m^2}} = \frac{7 \times 10^{-7} \mathrm{~T \cdot m / A} \cdot 0.60 \times 10^{-3} \mathrm{~m}}{1.44 \times 10^{-6} \mathrm{~m^2}}\] \[B \approx 2.92 \times 10^{-3} \mathrm{~T}\] So, the magnitude of the magnetic field inside the tungsten wire at a distance of \(0.60 \mathrm{~mm}\) from its central axis is approximately \(2.92 \times 10^{-3} \mathrm{~T}\).

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Most popular questions from this chapter

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius \(2.00 \cdot 10^{3} \mathrm{~km}\) through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about \(6.00 \cdot 10^{-5} \mathrm{~T}\). About how large a current would be required to produce such a field?

The number of turns in a solenoid is doubled, and its length is halved. How does its magnetic field change? a) it doubles b) it is halved c) it quadruples d) it remains unchanged

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

Many electrical applications use twisted-pair cables in which the ground and signal wires spiral about each other. Why?

A long solenoid (diameter of \(6.00 \mathrm{~cm}\) ) is wound with 1000 turns per meter of thin wire through which a current of 0.250 A is maintained. A wire carrying a current of 10.0 A is inserted along the axis of the solenoid. What is the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis?
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